Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long A[10][10] = {};
void rec(long long x, long long y, long long n) {
  if ((n % 2) == 0) {
    A[x][y] = n + 1;
  }
  if (n == 0)
    return;
  else {
    long long arrx[] = {-2, 2};
    long long arry[] = {-1, 1};
    for (auto i : arrx) {
      for (auto j : arry) {
        long long newx = x + i, newy = y + j;
        if (newx >= 0 && newx < 10 && newy >= 0 && newy < 10) {
          {
            cerr << "> "
                 << "newx"
                 << ": " << newx << " , "
                 << "newy"
                 << ": " << newy << endl;
          };
          if (A[newx][newy] < n) rec(newx, newy, n - 1);
        }
      }
    }
    for (auto j : arrx) {
      for (auto i : arry) {
        long long newx = x + i, newy = y + j;
        if (newx >= 0 && newx < 10 && newy >= 0 && newy < 10) {
          {
            cerr << "> "
                 << "newx"
                 << ": " << newx << " , "
                 << "newy"
                 << ": " << newy << endl;
          };
          if (A[newx][newy] < n) rec(newx, newy, n - 1);
        }
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    double n, h;
    cin >> n >> h;
    double f = h / sqrt(n);
    cout << setprecision(9);
    for (long long i = 0; i < n - 1; i++) {
      cout << f * sqrt(i + 1) << ' ';
    }
  }
}